Annual effective rate of interest

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September 8th 2009, 09:36 AM
tbl9301

Annual effective rate of interest

A bank offers the following certificates of deposit:
Term in years / Nominal annual interest rate compounded semi-annually
1 - 5%
2 - 6%
3 - 7%
4 - 8%
The bank does not permit early withdrawals. The certificates mature at the end of the term. During the next 6 years the bank will continue to offer these certificates of deposit. An investor initially deposits $10,000 in the bank and withdraws both principle and interest at the end of 6 years. Calculate the maximum annual effective rate of interest the investor can earn over the 6-year period. (The answer is given as 7.466%)

So I found the accumulated values for each to be, respectively: 10506.3 , 11225.1 , 12292.6 , 13685.7 ... but I'm not sure how to set it up to get the annual effective rates of interest.

Thank you!
September 8th 2009, 10:49 AM
aidan

Quote:

Originally Posted by tbl9301

A bank offers the following certificates of deposit:
Term in years / Nominal annual interest rate compounded semi-annually
1 - 5%
2 - 6%
3 - 7%
4 - 8%
The bank does not permit early withdrawals. The certificates mature at the end of the term. During the next 6 years the bank will continue to offer these certificates of deposit. An investor initially deposits $10,000 in the bank and withdraws both principle and interest at the end of 6 years. Calculate the maximum annual effective rate of interest the investor can earn over the 6-year period. (The answer is given as 7.466%)

So I found the accumulated values for each to be, respectively: 10506.3 , 11225.1 , 12292.6 , 13685.7 ... but I'm not sure how to set it up to get the annual effective rates of interest.

Thank you!

You must also determine which Option is best ?:
4 years @8% + 2 yrs @ 6%
or
2 years @6% + 4 years @8%
or
3 years @7% + 3 yrs at 7%
or
6 years @5%

Quote:

I found the accumulated values for each to be, respectively: 10506.3 , 11225.1 , 12292.6 , 13685.7

How did you get those numbers?
I ask because:
$(1.05)^{6} \times 10000 = 13400.96$
3400.96/6 = 566.83 average per year
&
5% compounded semiannually is 2.5% per 6 months.
$(1.025)^{12} \times 10000 = 13448.89$
Total interest earned is $3448.89 or $574.81 average per year.

I do not understand your 10506.3.
Could you explain what the 10506.3 value actually means?

.
September 8th 2009, 01:08 PM
tbl9301

Oh I was setting it up like, for the 5% for 1 year:
10000(1 + .05/2)2*1 = 10000(1.025)2 = 10506.30

Is that the wrong formula to use for this case?
September 8th 2009, 07:58 PM
aidan

Quote:

Originally Posted by tbl9301

Oh I was setting it up like, for the 5% for 1 year:
10000(1 + .05/2)2*1 = 10000(1.025)2 = 10506.30

Is that the wrong formula to use for this case?

That helps me understand the question.

5%:: $\left ( \left ( 1 + \dfrac{0 .05}{2} \right ) ^2 - 1 \right ) 100$ :: yields 5.0625 %

6%:: $\left( \left ( 1+\dfrac{0.06}{2} \right)^2 - 1 \right ) 100$ :: yields 6.0900%

7%:: $\left( \left ( 1+\dfrac{0.07}{2} \right)^2 - 1 \right ) 100$ :: yields 7.1225%

8%:: $\left ( \left ( 1+\dfrac{0.04}{2} \right)^2 - 1 \right ) 100$ :: yields 8.1600%

$\dfrac {4 ( 8.1600 ) + 2 ( 6.0900 )}{6} = \dfrac{44.82}{6} = 7.4700$
September 8th 2009, 11:46 PM
Wilmer

or (1.04^8 1.03^4)^(1/6) - 1 = .0746555.... ; ~7.47%

Amount invested does not affect calculation.